Function Notation for 'the bleebing of a number'

[KarlyD]

The Great Wizard of Gabadoo has described a method for what he calls "the bleebing of a number". He takes a real number, multiplies it by 3, then adds 4, and then subtracts one eighth of the result of multiplying the original number by (-2).

a) Write the recipe for "bleebing" using function notation.

b) What is the "anti-bleebing" formula (the inverse function of the bleebing function)?

My work:
a) (R x 3)+4 = x
[R(-2)] - (1/8)

Please help!! =(

 

[stapel]

I will guess that you are using "R" stand for "the real number" that you started with. But I'm not clear on how you're using "x"...?

Since you're supposed to create a function f(x), you might want to use "x" for the starting value, instead. Then do the translating that you learned in earlier algebra courses:

. . . . .starting number: x

Your function for "bleebing a number" will be f(x), a function f in terms of x. I think you already have some of the pieces:

. . . . ."three times" the number: 3x

. . . . ."adding four" to three times the number: 3x + 4

Then do the other pieces:

. . . . .multiplying the original number by -2: [fill this in]

. . . . .one-eighth of the result of the above: [fill this in]

. . . . .subtracting the last line above from 3x + 4: [fill this in]

Then simplify. This will give you f(x) = (the bleebing of x).

Then use whatever method they taught you for finding f^-1(x).

Eliz.

 

[KarlyD]

When I'm at 3x + 4, I don't understand how to show multiplying the original number by (-2). Is it just 3[(x)(-2)] + 4?

For one eighth of the result, is it (3[(x)(-2)] +4) / 8?

 

[stapel]

When I'm at 3x + 4, I don't understand how to show multiplying the original number by (-2). Is it just 3[(x)(-2)] + 4?

Why would you back up and start over again at the beginning, changing the first step...?

Instead, try following the instructions. You've already done the first bit:

. . . . .i) Pick a variable for "the number": x

. . . . .ii) Write an expression, in terms of (i), for
. . . . ."three times" the number: 3x

. . . . .iii) Write an expression, in terms of (ii), for
. . . . .adding four from (ii): 3x + 4

. . . . .iv) Take the variable for "the original number"
. . . . .from (i). Create an expression which shows
. . . . .the multiplication of "the number" by -2.

. . . . .v) Create an expression, in terms of (iv), for
. . . . .taking one-eighth of the value in (iv).

. . . . .vi) Subtract (v) from (iii).

. . . . .vii) Simplify.

Then find the inverse function for the function created in (vii).

Eliz.

 

[KarlyD]

I apologize for my ignorance, but I'm having a lot of trouble with this.



3x + 4
[3(-2x) + 4] = y
y-1/8=z

 

[stapel]

3x + 4

This is step (iii). So we're okay to there. But--

I'm afraid I'm not seeing where you tried steps (iv) through (vii), nor do I follow your reasoning for going back to the beginning and replacing "x" in step (i) with "-2x".

Please clarify your reasoning (for whatever method you're using), or else please try following the step-by-step instructions, starting with step (iv). :idea:

Thank you!

Eliz.

 

[KarlyD]

Sure, sorry!

x = "the number"

3x = three times the number

3x + 4 = three times the number plus 4

* I have to multiply the original number by -2 before subtracting 1/8th of the result

3(-2x) + 4 = three times the number, plus 4, multiplying the original number by 2

[3(-2x) + 4] - 1/8 = three times the number, plus 4, multiplying the original number by 2, subtracting 1/8th of the result

 

[stapel]

* I have to multiply the original number by -2 before subtracting 1/8th of the result

But does "the original number" indicate that you should "replace the variable in the last expression with something else, changing the meaning", or does it mean "go back to just 'x' and start creating a new expression"? :wink:

Eliz.

. . . . .iv) Take the variable for "the original number"
. . . . .from (i). Create an expression which shows
. . . . .the multiplication of "the number" by -2.

. . . . .v) Create an expression, in terms of (iv), for
. . . . .taking one-eighth of the value in (iv).

. . . . .vi) Subtract (v) from (iii).

. . . . .vii) Simplify.

Then find the inverse function for the function created in (vii).

 

[KarlyD]

I give up on this one.
:roll:

Thanks for your help!
:lol:

 

[stapel]

I give up on this one.

Why? If you can create an expression for "the number, multiplied by 3", you should be able to create an expression for "the number, multiplied by -2", right?

If you can create an expression for "add four to the last step", you should be able to create an expression for "take one-eighth of the last step".

Then subtract the second expression ("one-eighth of -2 times the number") from the first expression ("add four to three times the number"), and simplify.

I guess I'm not understanding why the second expression (steps (iv) through (vii)) is so different from the first expression (steps (i) through (iii))...?

Eliz.